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Calculus · Definite-Integrals

Integrate x·e^(-x) from 0 to oo

\int_{0}^{\infty} x e^{- x}\, dx

\int_{0}^{\infty} x e^{- x}\, dx
First find an antiderivative F, then evaluate F(b) − F(a).
u = - x,\quad du = -1\, dx
Substitute u = - x.
\int x e^{- x}\, dx = \int u e^{u}\, d_u
Rewrite the integral in terms of u.
u = u,\quad dv = e^{u}\, d_u
Integration by parts: ∫u dv = uv − ∫v du.
du = 1\, d_u,\quad v = e^{u}
Differentiate u, integrate dv.
\int u e^{u}\, d_u = u e^{u} - \int e^{u}\, d_u
Apply the formula.
\int e^{u}\, d_u = e^{u}
∫ aᵘ du = aᵘ / ln a (for eˣ that is just eˣ).
= u e^{u} - e^{u}
Combine.
= - x e^{- x} - e^{- x}
Substitute back u = - x.
F(\infty) - F(0) = \left(\text{NaN}\right) - \left(-1\right)
Fundamental theorem of calculus: plug in the limits.
= 1
Simplify.

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